In November 2024
Answer to Question #177265 in Abstract Algebra for Abhijeet
a) Prove that every non-trivial subgroup of a cyclic group has finite index. Hence
prove that (Q, +) is not cyclic. (7)
b) Let G be an infinite group such that for any non-trivial subgroup H of
G, G : H < ∞. Then prove that
i)
H ≤ G ⇒ H = {e} or H is infinite;
ii) If g ∈G, g ≠ e, then o(g) is infinite. (5)
c) Prove that a cyclic group with only one generator can have at most 2 elements. (3)
A)Given, G is a cylic group.
Therefore, G=<a>.
And H is subgroup of G.
H=<a^i>.
Index,
G/H={a^j+<a^i>}
If j>i then by division algorithm, j=ir+s
Then a^j+<a^i>=a^s+<a^i>
So, G/H={a^s+<a^i>, 0<=s<i}
O(G/H)= finite =index of H.
For( Q,+), choose H=<1/2>
Clearly o(G/H)= inifinte.
So, G=(Q,+) is not cyclic.
b)
Given, G be an infinite group such that for any non-trivial subgroup H of
G, G : H < ∞.
i)
If H is finite then order of G will be finite by Lagrange's theorem. Which is contradiction.
Thus, H = {e} or H is infinite.
ii)
let g be any element of group G.
G=<g>
Since, H is infinte using H=<1>.
Therefore, o(g)=infinte.
c)
Let a be an arbitrary element of cyclic group G.
Since, G is cyclic.
Therefore, it generated by its elements say x i.e., G=⟨x⟩.
G=⟨x⟩, then there exists an integer k such that a=xk=x2q+r, where r∈{0,1},
r is the remainder of the euclidean division and q is the quotient.
Then a=(x2)q.xr=1q.xr=xr, for r∈{0,1}. So, a can be 1 and x only.
Therefore, cyclic group G have atmost 2 elements.
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